Resistors in a circuit/ Calculate total resistance

[Rokas_P]

I'm having problem trying to calculate total resistance in resistor arrangement like this one:

http://o3xn.files.wordpress.com/2012/04/different.png

Before I go into any more detail with the original problem, I just want to say that whenever I get this "closed" type of circuits, I like rearranging them into the "open" type, so that e.g.

http://o3xn.files.wordpress.com/2012/04/r_1_20120414.png

becomes

http://o3xn.files.wordpress.com/2012/04/r_2_20120414.png

-------

Okay, back to the original question. First, I observe that potential at point 2 is higher than at point 3, which means that the current flows from 2 to 3 ("down"). This in turn means that there are basically two (parallel) ways for the current to pass this circuit:

(the above both) first R, then either 5R or 2R-7R / the two subpaths are in parallel to one another

(the below path) first 2R, then 7R

This seems to indicate that I can redraw the circuit as follows:

http://o3xn.files.wordpress.com/2012/04/different2.png

This rearranged circuit is not entirely "open" (2R is connected to the wire in between 2R and 7R) and I have no idea how I should treat this.

One of my guesses was that perhaps I am allowed to add another 7R resistor, making the circuit look like this one:

http://o3xn.files.wordpress.com/2012/04/different3.png

I'm not sure if I can add the extra resistor if it wasn't in the original problem. But it does have two different currents flowing, so maybe?

 

[gneill]

Your various alternate "paths" have influences on each other that you're not taking into account. When current flows through a resistor a potential drop ensues, and this potential drop has the effect of altering the currents through other paths. Thus they are all interrelated to such an extent that you must be very careful indeed when you make assumptions about what alterations are okay.

The best way to proceed with this kind of circuit is to either write KVL and/or KCL equations with an assumed voltage source across the circuit, or employ a Delta-Y transformation on one of the Δ shaped resistor groupings; That will allow you to proceed with the usual parallel/serial reductions.

 

[Rokas_P]

Okay, so this is what I get:

http://o3xn.files.wordpress.com/2012/04/r20120415_12.png

First, I slightly rearranged the circuit to clearly see the delta grouping in it, then I applied the Y-Δ transform, which eliminated node 2 and instead connected nodes 1-3 and 3-4. The equivalent resistors have the following resistances:

[itex]R_{\alpha\gamma}=\dfrac{R_\alpha R_\beta+R_\beta R_\gamma+R_\alpha R_\gamma}{R_\beta}=\dfrac{5R^2+10R^2+2R^2}{5R}= \dfrac{17}{5}R=3.4R[/itex]
[itex]R_{\alpha\beta}=\dfrac{R_\alpha R_\beta+R_\beta R_\gamma+R_\alpha R_\gamma}{R_\gamma}=\dfrac{5R^2+10R^2+2R^2}{2R}= \dfrac{17}{2}R=8.5R[/itex]
[itex]R_{\beta\gamma}=\dfrac{R_\alpha R_\beta+R_\beta R_\gamma+R_\alpha R_\gamma}{R_\alpha}=\dfrac{5R^2+10R^2+2R^2}{R}=17R[/itex]

Finally, I calculate the resistance to be:

[itex]R_{\alpha\gamma}\parallel2R=1\dfrac{7}{27}R[/itex]
[itex]R_{\beta\gamma}\parallel7R=4\dfrac{23}{24}R[/itex]

The two above in series gives us

[itex]R_\text{lower}=6\dfrac{47}{216}R[/itex]

Finally, resistance in the lower path in parallel with resistance in the upper path gives us

[itex]R_{\alpha\beta}\parallel R_\text{lower}=3\dfrac{221}{374}R[/itex]

Questions:

1. are the transformations correct (see the picture in the link)?
2. is the solution correct (and I mean more about the way I solve this, not the exact number)?

 

[gneill]

Your method is fine and your result is good, although the form ##\frac{79}{22}R## might be preferable for some

 

[Rokas_P]

Thanks!

I see that in all my excitement I didn't notice that I could simplify the fraction:

[itex]3\dfrac{221}{374}=3\dfrac{13}{22}[/itex]

Follow-up questions:

1. is it correct that in the original problem, potential is higher at 2 than at 3, which makes currwent flow from 2 to 3?

2. is it possible to solve this circuit without the complications of delta-y transformation or Kirchhoff's Laws (by simply determining which Rs are connected in series and which in parallel and determinig the direction of current flow)?

 

[gneill]

Thanks!

I see that in all my excitement I didn't notice that I could simplify the fraction:

[itex]3\dfrac{221}{374}=3\dfrac{13}{22}[/itex]

Follow-up questions:

1. is it correct that in the original problem, potential is higher at 2 than at 3, which makes currwent flow from 2 to 3?

Since no potential source is specified, it's not a quation that can be answered. If you impose a potential across nodes 1 and 4, the direction of the current will depend upon the direction of that potential.

2. is it possible to solve this circuit without the complications of delta-y transformation or Kirchhoff's Laws (by simply determining which Rs are connected in series and which in parallel and determinig the direction of current flow)?

Nope, since there are no candidate series or parallel connections to proceed.

 

[Rokas_P]

Normally I assume the potential to be higher at an input left and lower at an exit right, which gives the overall current direction from left to right. I realize that this may not always be the case but let's just make this assumption for now.

Let me present my reasoning now.

Once we assume [itex]\varphi_1>\varphi_4[/itex], we know the current flows from left to right. That means that for all paths in parallel that connect these two points, the drop in potential (thus the voltage) must be the same.

Now if we take, e.g. two serial resistors, R1 and R2 and know that voltage across both of them is U, in what proportions does it distribute between the two resistors?

Since current across elements in series is equal, we can write

[itex]I_1=I_2\quad\Rightarrow\quad \dfrac{U_1}{R_1}=\dfrac{U_2}{R_2}\quad\Rightarrow \quad \dfrac{U_1}{U_2}=\dfrac{R_1}{R_2}[/itex]

which gives us the very important conclusion that voltage across a chain of elements in series distributes proportionally to their resistances.

Let's denote [itex]\varphi_1-\varphi_4=U[/itex]. In our original problem, the current which takes the upper path and flows only across R and 5R must experience this potential drop. The voltage distributes across the two resistors so that

[itex] \dfrac{U_1}{U_2}=\dfrac{R}{5R}=\dfrac{1}{5}[/itex]

meaning that

[itex]U_1=\dfrac{1}{6}U;\quad U_2=\dfrac{5}{6}U[/itex]

and in a similar way (for the lower path, resistors 3 (2R) and 4 (7R),

[itex]U_3=\dfrac{2}{9}U;\quad U_4=\dfrac{7}{9}U[/itex].

Note that after the current has flowed across resistor 1, potential will have dropped by only [itex]\frac{1}{6}U=\frac{3}{18}U[/itex] whereas for resistor 3 it will have dropped by [itex]\frac{2}{9}U=\frac{4}{18}U[/itex]. So potential drop is greater once resistor 3 is passed as opposed to that when resistor 1 is passed, thus [itex]\varphi_2>\varphi_3[/itex]. Therefore, the current flows "downwards".

----

If we inverted input and output we could still reason in a similar fashion and would arrive at a non-contradictory conclusion.

----

I happen to often come across problems with similar circuit topology but where the resistors both to the left and to the right of the central wire have the same resistances. For example, in my original problem we could have had two 2R's on the left and then two 7R's on the right. This would have imparted certain symmetry to the problem and by the reasoning above we would have been able to show that current does not flow at all across the middle wire allowing us to dismiss it altogether.

I've seen such problems in first year undergraduate exams and was convinced that they are solved without too many complications. And they indeed are, as I've shown in this post.

However, once symmetry is removed the current flows in one of the two possible directions so the middle wire cannot be dismissed. Nevertheless, the direction of current flow in the central wire (given a potential difference is assumed between points 1 and 4) can be determined, so can the problem be solved with simple series-parallel reductions then?

 

[gneill]

Normally I assume the potential to be higher at an input left and lower at an exit right, which gives the overall current direction from left to right. I realize that this may not always be the case but let's just make this assumption for now.

Let me present my reasoning now.

Once we assume [itex]\varphi_1>\varphi_4[/itex], we know the current flows from left to right. That means that for all paths in parallel that connect these two points, the drop in potential (thus the voltage) must be the same.

Now if we take, e.g. two serial resistors, R1 and R2 and know that voltage across both of them is U, in what proportions does it distribute between the two resistors?

Since current across elements in series is equal, we can write

[itex]I_1=I_2\quad\Rightarrow\quad \dfrac{U_1}{R_1}=\dfrac{U_2}{R_2}\quad\Rightarrow \quad \dfrac{U_1}{U_2}=\dfrac{R_1}{R_2}[/itex]

which gives us the very important conclusion that voltage across a chain of elements in series distributes proportionally to their resistances.

Let's denote [itex]\varphi_1-\varphi_4=U[/itex]. In our original problem, the current which takes the upper path and flows only across R and 5R must experience this potential drop. The voltage distributes across the two resistors so that

[itex] \dfrac{U_1}{U_2}=\dfrac{R}{5R}=\dfrac{1}{5}[/itex]

meaning that

[itex]U_1=\dfrac{1}{6}U;\quad U_2=\dfrac{5}{6}U[/itex]

and in a similar way (for the lower path, resistors 3 (2R) and 4 (7R),

[itex]U_3=\dfrac{2}{9}U;\quad U_4=\dfrac{7}{9}U[/itex].

Unfortunately this statement (in red) does not hold. While it is true for resistors in series, the R and 5R resistors are NOT in series in this circuit: Their connection point is shared with another component (a 2R resistor).

This means that you cannot claim that the currents in both the R and 5R resistors are the same. Some unknown current will be added to or taken from the connection node via this other path, and in general you cannot tell off hand which of the two (or both) original resistors that current is going to be flowing through. Your ratio will not hold.

Note that after the current has flowed across resistor 1, potential will have dropped by only [itex]\frac{1}{6}U=\frac{3}{18}U[/itex] whereas for resistor 3 it will have dropped by [itex]\frac{2}{9}U=\frac{4}{18}U[/itex]. So potential drop is greater once resistor 3 is passed as opposed to that when resistor 1 is passed, thus [itex]\varphi_2>\varphi_3[/itex]. Therefore, the current flows "downwards".

----

If we inverted input and output we could still reason in a similar fashion and would arrive at a non-contradictory conclusion.

----

I happen to often come across problems with similar circuit topology but where the resistors both to the left and to the right of the central wire have the same resistances. For example, in my original problem we could have had two 2R's on the left and then two 7R's on the right. This would have imparted certain symmetry to the problem and by the reasoning above we would have been able to show that current does not flow at all across the middle wire allowing us to dismiss it altogether.

The symmetry is purposely included in the problem in order to make it simpler to analyze and to sensitize the student to symmetry as a powerful simplifying tool.

I've seen such problems in first year undergraduate exams and was convinced that they are solved without too many complications. And they indeed are, as I've shown in this post.

However, once symmetry is removed the current flows in one of the two possible directions so the middle wire cannot be dismissed. Nevertheless, the direction of current flow in the central wire (given a potential difference is assumed between points 1 and 4) can be determined, so can the problem be solved with simple series-parallel reductions then?

Nope.

 

[Rokas_P]

Okay, thanks. We can consider this issue fully discussed.

 

[Rokas_P]

OK, now I'm trying to solve it using KCL/KVL and expect the same answer. I don't see, however, in what way KCL/KVL should help me determine how the resistors are connected.

First, I redraw the circuit and choose arbitrary currents and loop directions. I also apply a dummy potential U₀ across the circuit.

http://s18.postimage.org/6r3og1715/Drawing1.png

Then I write down equations for KCL (how many do I need?):

[itex]
i_1-i_2+i_4=0\\
i_2-i_3-i_5=0\\
i_3-i_4+i_6=0\\
i_5-i_6-i_1=0[/itex]

and KVL (how many?):

[itex]
-U_0+i_4\cdot2R+i_6\cdot7R=0\\
-i_2\cdot R-i_3\cdot2R-i_4\cdot2R=0\\
-i_5\cdot5R-i_6\cdot7R+i_3\cdot2R=0[/itex]

My guess is that I need the true directions of all the currents (6 of them), then I would know how the resistors are connected. For that, I need 6 equations and solve for i₁-i₆.

I cannot get any numerical value from this system but I can at least get the signs and determine the direction of the currents.

When I solve for the currents, I get

[itex]
i_1=-\dfrac{5U_0}{41R}\\
i_2=-\dfrac{2U_0}{41R}\\
i_3=-\dfrac{2U_0}{41R}\\
i_4=\dfrac{3U_0}{41R}\\
i_5=0\\
i_6=\dfrac{5U_0}{41R}[/itex]

This, I believe, is wrong.

 

[gneill]

The fundamental statement of KCL is that the sum of currents entering and leaving a node is zero. This fact allows you to write a minimal set of KCL equations if you only add "new" currents to your schematic when necessary as you step through the circuit and label them. The "true" directions of currents will sort themselves out in the math, so don't spend a lot of time worrying about it at the outset. I find it simple to just begin with the source and label it i1 (as you've done) and then proceed through the circuit adding new currents only as required when a path decision (split) must be made.

So for example, if you start with your i1 from the source and proceed to node 1 you see that a split is required. So you label one branch with a new current, i2, as you've done, but the other you can then label with "i1 - i2" rather than "i4", and its direction would then be downward, flowing out of node 1.

Similarly for node 2 there's a split required, so you can stick an i3 going off in one direction as you've done, and then the current left to flow through the 5R resistor is "i2 - i3". Now there's only one component that's still lacking a current, and that's the 7R resistor. At node 3 we have currents "i1 - 12" and "i3" flowing in, and only one path out through the 7R, so the current in that resistor can be labelled "i1 - i2 + i3".

Now you're in a position to write some KVL loop equations with those currents for the components. You need just enough loops so that every component is included in at least one of them. So, three loops will suffice in this case.

If you solve for i1, then Req = Uo/i1 .

 

[Rokas_P]

Your advice really simplifies things. Thanks!

I did as you suggested, and solved for i1 (actually I just simplified the system of the three equations and then handed it off for MathCAD to handle - there's too much physics to be bothered by maths here). I got

[itex]i_1=-\dfrac{22U_0}{79R}[/itex],

and finally,

[itex]R_\text{eq}=\dfrac{U_0}{\left|i_1\right|}=\dfrac{79}{22}R[/itex],

which matches the answer from the Y-Δ transformation method.

(I believe the absolute value sign is necessary to accommodate for the fact that the initial choice for the current direction may be wrong)

Luckily i1 contained a U0 in it so that the U0 from the Req's definition could cancel out. I assume that's always the case when you apply a fake potential across a circuit you're solving?

P.S. at node 3, I first had a situation where all currents were flowing into node 3 (since that was how I instinctively placed current directions as I was making my way through the circuit). I felt this was wrong, so I changed the direction of the current between nodes 3 and 4 into the opposite. But this feels sort of wrong. Shouldn't I be allowed to choose my directions however I want and still get the right answer as loong as I am careful with the signs?

 

[gneill]

Your advice really simplifies things. Thanks!

I did as you suggested, and solved for i1 (actually I just simplified the system of the three equations and then handed it off for MathCAD to handle - there's too much physics to be bothered by maths here). I got

[itex]i_1=-\dfrac{22U_0}{79R}[/itex],

and finally,

[itex]R_\text{eq}=\dfrac{U_0}{\left|i_1\right|}=\dfrac{79}{22}R[/itex],

which matches the answer from the Y-Δ transformation method.

(I believe the absolute value sign is necessary to accommodate for the fact that the initial choice for the current direction may be wrong)

Actually, as long as you're consistent with the signs of sources and directions of currents (thus directions of potential changes), the result should be a positive value. The circuit resistance will be given by the source voltage divided by the current flowing OUT of the + terminal of that source.

Luckily i1 contained a U0 in it so that the U0 from the Req's definition could cancel out. I assume that's always the case when you apply a fake potential across a circuit you're solving?

Yup.

P.S. at node 3, I first had a situation where all currents were flowing into node 3 (since that was how I instinctively placed current directions as I was making my way through the circuit). I felt this was wrong, so I changed the direction of the current between nodes 3 and 4 into the opposite. But this feels sort of wrong. Shouldn't I be allowed to choose my directions however I want and still get the right answer as loong as I am careful with the signs?

Yes. The math should always take care of itself. Arbitrary decisions about current direction are fine so long as they are made only once